Wave Optics
Chapter 10
WAVE OPTICS
MCQ I
10.1 Consider a light beam incident from air to a glass slab at Brewster’s angle as shown in Fig. 10.1.
A polaroid is placed in the path of the emergent ray at point $P$ and rotated about an axis passing through the centre and perpendicular to the plane of the polaroid.
(a) For a particular orientation there shall be darkness as observed through the polaoid.
(b) The intensity of light as seen through the polaroid shall be independent of the rotation.
(c) The intensity of light as seen through the Polaroid shall go through a minimum but not zero for two orientations of the polaroid.
(d) The intensity of light as seen through the polaroid shall go through a minimum for four orientations of the polaroid.
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Answer: (c)10.2 Consider sunlight incident on a slit of width $10^{4} A$. The image seen through the slit shall (a) be a fine sharp slit white in colour at the center.
(b) a bright slit white at the center diffusing to zero intensities at the edges.
(c) a bright slit white at the center diffusing to regions of different colours.
(d) only be a diffused slit white in colour.
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Answer: (a)10.3 Consider a ray of light incident from air onto a slab of glass (refractive index $n$ ) of width $d$, at an angle $\theta$. The phase difference between the ray reflected by the top surface of the glass and the bottom surface is
(a) $\frac{4 \pi d}{\lambda}(1-\frac{1}{n^{2}} \sin ^{2} \theta)^{1 / 2}+\pi$
(b) $\frac{4 \pi d}{\lambda}(1-\frac{1}{n^{2}} \sin ^{2} \theta)^{1 / 2}$
(c) $\frac{4 \pi d}{\lambda}(1-\frac{1}{n^{2}} \sin ^{2} \theta)^{1 / 2}+\frac{\pi}{2}$
(d) $\frac{4 \pi d}{\lambda}(1-\frac{1}{n^{2}} \sin ^{2} \theta)^{1 / 2}+2 \pi$.
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Answer: (a)10.4 In a Young’s double slit experiment, the source is white light. One of the holes is covered by a red filter and another by a blue filter. In this case
(a) there shall be alternate interference patterns of red and blue.
(b) there shall be an interference pattern for red distinct from that for blue.
(c) there shall be no interference fringes.
(d) there shall be an interference pattern for red mixing with one for blue.
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Answer: (c)10.5 Figure 10.2 shows a standard two slit arrangement with slits $S_1, S_2 . P_1, P_2$ are the two minima points on either side of $P$ (Fig. 10.2).
At $P_2$ on the screen, there is a hole and behind $P_2$ is a second 2- slit arrangement with slits $S_3, S_4$ and a second screen behind them.
Fig. 10.2
(a) There would be no interference pattern on the second screen but it would be lighted.
(b) The second screen would be totally dark.
(c) There would be a single bright point on the second screen.
(d) There would be a regular two slit pattern on the second screen.
MCQ II
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Answer: (d)10.6 Two source $S_1$ and $S_2$ of intensity $I_1$ and $I_2$ are placed in front of a screen [Fig. 10.3 (a)]. The patteren of intensity distribution seen in the central portion is given by Fig. 10.3 (b).
In this case which of the following statements are true.
Fig. 10.3 (a)
(a) $S_1$ and $S_2$ have the same intensities.
(b) $S_1$ and $S_2$ have a constant phase difference.
(c) $S_1$ and $S_2$ have the same phase.
(d) $S_1$ and $S_2$ have the same wavelength.
Fig. 10.3 (b)
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Answer: (a), (b), (d)10.7 Consider sunlight incident on a pinhole of width $10^{3} A$. The image of the pinhole seen on a screen shall be
(a) a sharp white ring.
(b) different from a geometrical image.
(c) a diffused central spot, white in colour.
(d) diffused coloured region around a sharp central white spot.
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Answer: (b), (d)10.8 Consider the diffraction patern for a small pinhole. As the size of the hole is increased
(a) the size decreases.
(b) the intensity increases.
(c) the size increases.
(d) the intensity decreases.
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Answer: (a), (b)10.9 For light diverging from a point source
(a) the wavefront is spherical.
(b) the intensity decreases in proportion to the distance squared.
(c) the wavefront is parabolic.
(d) the intensity at the wavefront does not depend on the distance.
VSA
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Answer: (a), (b)10.10 Is Huygen’s principle valid for longitudunal sound waves?
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Answer: Yes.10.11 Consider a point at the focal point of a convergent lens. Another convergent lens of short focal length is placed on the other side. What is the nature of the wavefronts emerging from the final image?
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Answer: Spherical.10.12 What is the shape of the wavefront on earth for sunlight?
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Answer: Spherical with huge radius as compared to the earth’s radius so that it is almost a plane.10.13 Why is the diffraction of sound waves more evident in daily experience than that of light wave?
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Answer: Sound wave have frequencies $20 Hz$ to $20 kHz$. The corresponding wavelengths are $15 m$ and $15 mm$, respectively. Diffraction effects are seen if there are slits of width $a$ such that.
$\alpha \sim \lambda$
For light waves, wavelengths $\sim 10^{-7} m$. Thus diffraction effects will show when
$a \sim 10^{-7} m$.
whereas for sound they will show for
$15 mm<a<15 m$.
Exemplar Problems-Physics
10.14 The human eye has an approximate angular resolution of $\phi=5.8 \times 10^{-4} rad$ and a typical photoprinter prints a minimum of $300 dpi$ (dots per inch, $1 inch=2.54 cm$ ). At what minimal distance $z$ should a printed page be held so that one does not see the individual dots.
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Answer: The linear distance between two dots is $l=\frac{2.54}{300} cm \simeq 0.84 \times 10^{-2} cm$.
At a distance of $Z cm$ this subtends an angle.
$\phi \sim l / z \therefore z=\frac{l}{\phi}=\frac{0.84 \times 10^{-2} cm}{5.8 \times 10^{-4}} \sim 14.5 cm$.
10.15 A polariod (I) is placed in front of a monochromatic source. Another polatiod (II) is placed in front of this polaroid (I) and rotated till no light passes. A third polaroid (III) is now placed in between (I) and (II). In this case, will light emerge from (II). Explain.
SA
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Answer: Only in the special cases when the pass axis of (III) is parollel to (I) or (II) there shall be no light emerging. In all other cases there shall be light emerging because the pass axis of (II) is no longer perpendicular to the pass axis of (III).10.16 Can reflection result in plane polarised light if the light is incident on the interface from the side with higher refractive index?
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Answer: Polarisation by reflection occurs when the angle of incidence is the Brewster’s angle i.e. $\tan \theta_B=\frac{n_2}{n_1}$ where $n_2<n_1$.
When light travels in such a medium the critical angle is $\sin \theta_c=\frac{n_2}{n_1}$ where $n_2<n_1$.
As $|\tan \theta_B|>|\sin \theta_c|$ for large angles, $\theta_B<\theta_C$.
Thus, polarisation by reflection shall definitely occur.
10.17 For the same objective, find the ratio of the least separation between two points to be distinguished by a microscope for light of $5000 \AA$ and electrons accelerated through $100 V$ used as the illuminating substance.
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Answer: $d _{\min }=\frac{1.22 \lambda}{2 \sin \beta}$
where $\beta$ is the angle subtended by the objective at the object.
For light of $5500 \AA$
$d _{\min }=\frac{1.22 \times 5.5 \times 10^{-7}}{2 \sin \beta} m$
For electrons accelerated through $100 V$ the deBroglie wavelength is
$\lambda=\frac{h}{p}=\frac{1.227}{\sqrt{100}}=0.13 nm=0.13 \times 10^{-9} m$
$\therefore d _{\text{min }}^{\prime}=\frac{1.22 \times 1.3 \times 10^{-10}}{2 \sin \beta}$ $\therefore d _{\min }^{\prime}=\frac{1.22 \times 1.3 \times 10^{-10}}{2 \sin \beta}$
$\frac{d _{\min }^{\prime}}{d _{\min }}=\frac{1.3 \times 10^{-10}}{5.5 \times 10^{-7}} \sim 0.2 \times 10^{-3}$
10.18 Consider a two slit interference arrangements (Fig. 10.4) such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of $D$ in terms of $\lambda$ such that the first minima on the screen falls at a distance $D$ from the centre $O$.
Fig. 10.4
LA
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Answer: $T_2 P=D+x, T_1 P=D-x$
$S_1 P=\sqrt{(S_1 T_1)^{2}+(PT_1)^{2}}$
$ =[D^{2}+(D-x)^{2}]^{1 / 2} $
$S_2 P=[D^{2}+(D+x)^{2}]^{1 / 2}$
Minima will occur when
$[D^{2}+(D+x)^{2}]^{1 / 2}-[D^{2}+(D-x)^{2}]^{1 / 2}=\frac{\lambda}{2}$
If $x=D$
$(D^{2}+4 D^{2})^{1 / 2}=\frac{\lambda}{2}$
$(5 D^{2})^{1 / 2}=\frac{\lambda}{2}, \quad \therefore D=\frac{\lambda}{2 \sqrt{5}}$.
10.19 Figure 10.5 shown a two slit arrangement with a source which emits unpolarised light. $P$ is a polariser with axis whose direction is not given. If $I_o$ is the intensity of the principal maxima when no polariser is present, calculate in the present case, the intensity of the principal maxima as well as of the first minima.
Fig. 10.5
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Answer: Without P:
$A=A _{\perp}+A _{11}$
$A _{\perp}=A _{\perp}^{1}+A _{\perp}^{2}=A _{\perp}^{0} \sin (k x-\omega t)+A _{\perp}^{0} \sin (k x-\omega t+\phi)$
$A _{11}=A _11^{(1)}+A _11^{(2)}$
$A _{11}=A _11^{0}[\sin (k x-\omega t)+\sin (k x-\omega t+\phi]$
where $A _{\perp}^{0}, A _11^{0}$ are the amplitudes of either of the beam in $\perp$ and 11 polarizations.
$\therefore$ Intensity $=$
$={|A _{\perp}^{0}|^{2}+|A _11^{0}|^{2}}[\sin ^{2}(k x-w t)(1+\cos ^{2} \phi+2 \sin \phi)+\sin ^{2}(k x-\omega t) \sin ^{2} \phi] _{\text{average }}$
$={|A _{\perp}^{0}|^{2}+|A _11^{0}|^{2}}(\frac{1}{2}) \cdot 2(1+\cos \phi)$
Exemplar Problems-Physics
$ =2|A _{\perp}^{0}|^{2} \cdot(1+\cos \phi) \text{ since }|A _{\perp}^{0}| _{\text{average }}=|A _11^{0}| _{\text{average }} $
With P:
Assume $A _{\perp}^{2}$ is blocked:
Intensity $=(A _11^{1}+A _11^{2})^{2}+(A _{\perp}^{1})^{2}$
$=|A _{\perp}^{0}|^{2}(1+\cos \phi)+|A _{\perp}^{0}|^{2} \cdot \frac{1}{2}$
Given: $I_0=4|A _{\perp}^{0}|^{2}=$ Intensity without polariser at principal maxima.
Intensity at principal maxima with polariser
$ \begin{aligned} & =|A _{\perp}^{0}|^{2}(2+\frac{1}{2}) \\ & =\frac{5}{8} I_0 \end{aligned} $
Intensity at first minima with polariser
$ \begin{aligned} & =|A _{\perp}^{0}|^{2}(1-1)+\frac{|A _{\perp}^{0}|^{2}}{2} \\ & =\frac{I_0}{8} . \end{aligned} $
10.20
Fig. 10.6
$AC=CO=D, S_1 C=S_2 C=d«D$
A small transparent slab containing material of $\mu=1.5$ is placed along $AS_2$ (Fig. 10.6). What will be the distance from $O$ of the principal maxima and of the first minima on either side of the principal maxima obtained in the absence of the glass slab. .
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Answer: Path difference $=2 d \sin \theta+(\mu-1) l$
$\therefore$ For principal maxima,
$2 d \sin \theta+0.5 l=0$
$\sin \theta_0=\frac{-l}{4 d}=\frac{-1}{16} \quad(\because l=\frac{d}{4})$
$\therefore OP=D \tan \theta_0 \approx-\frac{D}{16}$
For the first minima:
$\therefore 2 d \sin \theta_1+0.5 l= \pm \frac{\lambda}{2}$ $\sin \theta_1=\frac{ \pm \lambda / 2-0.5 l}{2 d}=\frac{ \pm \lambda / 2-\lambda / 8}{2 \lambda}= \pm \frac{1}{4}-\frac{1}{16}$
On the positive side: $\sin \theta^{+}=\frac{3}{16}$
On the negative side: $\sin \theta^{-}=-\frac{5}{16}$
The first principal maxima on the positive side is at distance
$D \tan \theta^{+}=D \frac{\sin \theta^{+}}{\sqrt{1-\sin ^{2} \theta}}=D \frac{3}{\sqrt{16^{2}-3^{2}}}$ above O.
In the -ve side, the distance will be $D \tan \theta^{-}=\frac{5}{\sqrt{16^{2}-5^{2}}}$ below $O$.
10.21 Four identical monochromatic sources A,B,C,D as shown in the (Fig.10.7) produce waves of the same wavelength $\lambda$ and are coherent. Two receiver $R_1$ and $R_2$ are at great but equal distaces from $B$.
(i) Which of the two receivers picks up the larger signal?
(ii) Which of the two receivers picks up the larger signal when B is turned off?
(iii) Which of the two receivers picks up the larger signal when D is turned off?
(iv) Which of the two receivers can distinguish which of the sources $B$ or $D$ has been turned off?
Fig. 10.7
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Answer: (i) Consider the disturbances at $R_1$ which is a distance $d$ from $A$. Let the wave at $R_1$ because of A be $Y_A=a \cos \omega t$. The path difference of the signal from $A$ with that from $B$ is $\lambda / 2$ and hence the phase difference is $\pi$.
Thus the wave at $R_1$ because of $B$ is
$y_B=a \cos (\omega t-\pi)=-a \cos \omega t$.
The path difference of the signal from $C$ with that from $A$ is $\lambda$ and hence the phase difference is $2 \pi$.
Thus the wave at $R_1$ because of $C$ is $y_c=a \cos \omega t$.
The path difference between the signal from $D$ with that of $A$ is
$\sqrt{d^{2}+(\frac{\lambda}{2})^{2}}-(d-\lambda / 2)$
$=d(1+\frac{\lambda}{4 d^{2}})^{1 / 2}-d+\frac{\lambda}{2}$
$=d(1+\frac{\lambda^{2}}{8 d^{2}})^{1 / 2}-d+\frac{\lambda}{2}$
If $d»\lambda$ the path difference $\sim \frac{\lambda}{2}$ and hence the phase difference is $\pi$.
Exemplar Problems-Physics
$\therefore y_D=-a \cos \omega t$
Thus, the signal picked up at $R_1$ is
$y_A+y_B+y_C+y_D=0$
Let the signal picked up at $R_2$ from $B$ be $y_B=a_1 \cos \omega t$.
The path difference between signal at $D$ and that at $B$ is $\lambda / 2$.
$\therefore y_D=-a_1 \cos \omega t$
The path difference between signal at $A$ and that at $B$ is
$\sqrt{(d)^{2}+(\frac{\lambda}{2})^{2}}-d=d(1+\frac{\lambda^{2}}{4 d^{2}})^{1 / 2}-d \sim \frac{1}{8} \frac{\lambda^{2}}{d^{2}}$
$\therefore$ The phase difference is $\frac{2 \pi}{8 \lambda} \cdot \frac{\lambda^{2}}{d^{2}}=\frac{\pi \lambda}{4 d}=\phi \sim 0$.
Hence, $y_A=a_1 \cos (\omega t-\phi)$
Similarly, $y_C=a_1 \cos (\omega t-\phi)$
$\therefore$ Signal picked up by $R_2$ is
$y_A+y_B+y_C+y_D=y=2 a_1 \cos (\omega t-\phi)$
$\therefore|y|^{2}=4 a_1^{2} \cos ^{2}(\omega t-\phi)$
$\therefore\langle I\rangle=2 a_1^{2}$
Thus $R_1$ picks up the larger signal.
(ii) If $B$ is switched off,
$R_1$ picks up $y=a \cos \omega t$
$\therefore\langle I _{R_1}\rangle=\frac{1}{2} a^{2}$
$R_2$ picks up $y=a \cos \omega t$
$\therefore\langle I _{R_2}\rangle=\frac{1}{2} a_1^{2}$
Thus $R_1$ and $R_2$ pick up the same signal.
(c) If D is switched off.
$R_1$ picks up $y=a \cos \omega t$
$\therefore\langle I _{R_1}\rangle=\frac{1}{2} a^{2}$
$R_2$ picks up $y=3 a \cos \omega t$
$\therefore\langle I _{R_2}\rangle=\frac{1}{2} 9 a^{2}$
Thus $R_2$ picks up larger signal compared to $R_1$.
(iv) Thus a signal at $R_1$ indicates $B$ has been switched off and an enhanced signal at $R_2$ indicates $D$ has been switched off.
10.22 The optical properties of a medium are governed by the relative permitivity $(\varepsilon_r)$ and relative permeability $(\mu_r)$. The refractive index is defined as $\sqrt{\mu_r \varepsilon_r}=n$. For ordinary material $\varepsilon_r>0$ and $\mu_r>0$ and the positive sign is taken for the square root. In 1964, a Russian scientist $V$. Veselago postulated the existence of material with $\varepsilon_r$ $<0$ and $\mu_r<0$. Since then such ‘metamaterials’ have been produced in the laboratories and their optical properties studied.
For such materials $n=-\sqrt{\mu_r \varepsilon_r}$. As light enters a medium of such refractive index the phases travel away from the direction of propagation.
(i) According to the description above show that if rays of light enter such a medium from air (refractive index $=1$ ) at an angle $\theta$ in $2^{\text{nd }}$ quadrant, them the refracted beam is in the $3^{\text{rd }}$ quadrant.
(ii) Prove that Snell’s law holds for such a medium.
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Answer: (i) Suppose the postulate is true, then two parallel rays would proceed as shown in Fig. 1. Assuming ED shows a wave front then all points on this must have the same phase. All points with the same optical path length must have the same phase.
Fig. 1
Fig. 2 Thus $-\sqrt{\varepsilon_r \mu_r} A E=BC-\sqrt{\varepsilon_r \mu_r} CD$
or $B C=\sqrt{\varepsilon_r \mu_r}(C D-A E)$
$BC>0, CD>AE$
As showing that the postulate is reasonable. If however, the light proceeded in the sense it does for ordinary material (viz. in the fourth quadrant, Fig. 2)
Then $-\sqrt{\varepsilon_r \mu_r} A E=BC-\sqrt{\varepsilon_r \mu_r} CD$
or, $B C=\sqrt{\varepsilon_r \mu_r}(C D-A E)$
As $AE>CD, BC<O$
showing that this is not possible. Hence the postalate is correct.
(ii) From Fig. 1.
$BC=AC \sin \theta_i$ and $CD-AE=AC \sin \theta_r:$
$ \begin{gathered} \text{ Since }-\sqrt{\varepsilon_r \mu_r}(A E-C D)=B C \\ -n \sin \theta_r=\sin \theta_i . \end{gathered} $
10.23 To ensure almost 100 per cent transmittivity, photographic lenses are often coated with a thin layer of dielectric material. The refractive index of this material is intermediated between that of air and glass (which makes the optical element of the lens). A typically used dielectric film is $MgF_2(n=1.38)$. What should the thickness of the film be so that at the center of the visible speetrum ( $5500 \AA$ ) there is maximum transmission.
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Answer: Consider a ray incident at an angle $i$. A part of this ray is reflected from the air-film interface and a part refracted inside. This is partly reflected at the film-glass interface and a part transmitted. A part of the reflected ray is reflected at the film-air interface and a part transmitted as $r_2$ parallel to $r_1$. Of course succesive reflections and transmissions will keep on decreasing the amplitude of the wave. Hence rays $r_1$ and $r_2$ shall dominate the behavior. If incident light is to be transmitted through the lens, $r_1$ and $r_2$ should interfere destructively. Both the reflections at $A$ and $D$ are from lower to higher refractive index and hence there is no phase change on reflection. The optical path difference between $r_2$ and $r_1$ is $n(AD+CD)-AB$.
If $d$ is the thickness of the film, then
$ A D=C D=\frac{d}{\cos r} $
$A B=A C \sin i$
$\frac{A C}{2}=d \tan r$
$\therefore A C=2 d \tan r$
Hence, $AB=2 d \tan r \sin i$
Thus the optical path difference is
$2 n \frac{d}{\cos r}-2 d \tan r \sin i$
$=2 \cdot \frac{\sin i}{\sin r} \frac{d}{\cos r}-2 d \frac{\sin r}{\cos r} \sin i$
$=2 d \sin [\frac{1-\sin ^{2} r}{\sin r \cos r}]$
$=2 n d \cos r$
For these waves to interefere destructively this must be $\lambda / 2$.
$\Rightarrow 2 n d \cos r=\frac{\lambda}{2}$
or $n d \cos r=\lambda / 4$
For a camera lens, the sources are in the vertical plane and hence $i \simeq r \simeq 0$
$\therefore n d \simeq \frac{\lambda}{4}$.
$\Rightarrow d=\frac{5500 \AA}{1.38 \times 4} \simeq 1000 \AA$
